CF 426C--Sereja and Swaps

As usual, Sereja has array a, its elements are integers: a[1], a[2], ..., a[n].
Let's introduce notation:



A swap operation is the following sequence of actions:

choose two indexes i, j (i ≠ j);

perform assignments tmp = a[i], a[i] = a[j], a[j] = tmp.

What maximum value of function m(a) can Sereja get if he is allowed
to perform at most k swap operations?

Input

The first line contains two integers n and k (1 ≤ n ≤ 200; 1 ≤ k ≤ 10).
The next line contains n integers a[1], a[2], ..., a[n] ( - 1000 ≤ a[i] ≤ 1000).

Output

In a single line print the maximum value of m(a) that Sereja can get
if he is allowed to perform at most k swap operations.

Sample test(s)

input

10 2
10 -1 2 2 2 2 2 2 -1 10

output

32

input

5 10
-1 -1 -1 -1 -1

output

-1

思路:dp[i][j][k]表示i到j这一段与其他进行K次操作所能达到的最大和，只需给i到j这一段建个最小堆，外面那部分建个最大堆，每次从最小堆里拿一个和最大堆的交换即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define inf 0x3f3f3f3f
#define maxn 220
int a[maxn];
int sum[maxn];
int dp[maxn][maxn][12];
inline int min(int a,int b)
{
return a>b?b:a;
}
inline int max(int a,int b)
{
return a>b?a:b;
}
struct Min
{
int aa;
Min(){}
Min(int x)
{
aa = x;
}
bool operator < (const Min & a)const
{
return aa > a.aa;
}
};
struct Max
{
int aa;
Max(){}
Max(int x)
{
aa = x;
}
bool operator < (const Max & a)const
{
return aa < a.aa;
}
};
priority_queue <Min> qmin;
priority_queue <Max> qmax;
int main()
{
//freopen("in.txt","r",stdin);
int n,K;
while(scanf("%d%d",&n,&K)==2)
{
for(int i = 1;i <= n;i++)
scanf("%d",&a[i]);
sum[0] = 0;
for(int i = 1;i <= n;i++)
sum[i] = sum[i-1] + a[i];

for(int i = 1;i <= n;i++)
{
for(int j = i;j <= n;j++)
{
int fuck = sum[j] - sum[i-1];
//这个就是区间的和。
while(!qmin.empty())    qmin.pop();
while(!qmax.empty())    qmax.pop();
for(int k = i;k <= j;k++)
{
qmin.push(Min(a[k]));
}
for(int k = 1;k < i;k++)
{
qmax.push(Max(a[k]));
}
for(int k = j+1;k <= n;k++)
{
qmax.push(Max(a[k]));
}
for(int k = 1;k <= K;k++)
{
int aa = 0,bb = 0;
if((!qmin.empty()) && (!qmax.empty()))
{
aa = qmin.top().aa;
bb = qmax.top().aa;
qmin.pop(); qmax.pop();
fuck += bb - aa;
qmin.push(Min(bb));
qmax.push(Max(aa));
}
dp[i][j][k] = fuck;
}
}
}
int ans = -inf;
for(int i = 1;i <= n;i++)
for(int j = i;j <= n;j++)
for(int k = 1;k <= K;k++)
{
ans = max(ans,sum[j] - sum[i-1]);
ans = max(ans,dp[i][j][k]);
}
printf("%d\n",ans);
}
return 0;
}